How many moles of bromide ions are present in 750.0 mL of 1.35 M MgBr₂?

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Firstly, we need to determine how many moles of MgBr₂ are present in 750.0 mL of 1.35 M solution: moles of MgBr₂ = (volume of solution in liters) x (molarity) moles of MgBr₂ = 0.750 L x 1.35 mol/L moles of MgBr₂ = 1.01 mol Next, we need to determine the number of bromide ions present for every one formula unit of MgBr₂. From the formula, we can see that there is one Mg2+ ion and two Br- ions per formula unit. Therefore: moles of Br- ions = (moles of MgBr₂) x 2 moles of Br- ions = 1.01 mol x 2 moles of Br- ions = 2.03 mol Therefore, the answer is C) 2.03 mol.

How Many Moles of Bromide Ions Are Present in 750